Let $I = \int_{ - \pi /4}^{\pi /4} {\log } (\sin x + \cos x)dx$ ……(i)



$I = \int_{ - \pi /4}^{\pi /4} {\log } \left\{ {\sin \left( {\frac{\pi }{4} - \frac{\pi }{4} - x} \right) + \cos \left( {\frac{\pi }{4} - \frac{\pi }{4} - x} \right)} \right\}dx$

$ = \int_{ - \pi /4}^{\pi /4} {\log } \{ \sin ( - x) + \cos ( - x)\} dx$

and $I = \int_{ - \pi /4}^{\pi 4} {\log } (\cos x - \sin x)dx$ ……(ii)

From Eqs. (i) and (ii),
$2I = \int_{ - \pi /4}^{\pi 4} {\log } \cos 2xdx$
$2I = \int_0^{\pi /4} {\log } \cos 2xdx$ …..(iii)
, if $\left. {f( - x) = f(x)} \right]$

Let's put $2x = t \Rightarrow dx = \frac{{dt}}{2}$

As $x \to 0$, then $t \to 0$

and $x \to \frac{\pi }{4}$, then $t \to \frac{\pi }{2}$

$2I = \frac{1}{2}\int_0^{\pi /2} {\log } \cos tdt$ …..(iv)

$ \Rightarrow $$2I = \frac{1}{2}\int_0^{\pi /2} {\log } \sin tdx$ …….(v)

On adding Eqs. (iv) and (v), we get
$4I = \frac{1}{2}\int_0^{\pi /2} {\log } \sin t\cos tdt$

$ \Rightarrow $$4I = \frac{1}{2}\int_0^{\pi /2} {\log } \frac{{\sin 2t}}{2}dt$

$ \Rightarrow $$4I = \frac{1}{2}\int_0^{\pi /2} {\log } \sin 2xdx - \frac{1}{2}\int_0^{\pi /2} {\log } 2dx$
$ \Rightarrow $$4I = \frac{1}{2}\int_0^{\pi /2} {\log } \sin \left( {\frac{\pi }{2} - 2x} \right)dx - \log 2 \cdot \frac{\pi }{4}$

$ \Rightarrow $$4I = \frac{1}{2}\int_0^{\pi /2} {\log } \cos 2xdx - \frac{\pi }{4}\log 2$

$ \Rightarrow $$4I = \int_0^{\pi /4} {\log } \cos 2xdx - \frac{\pi }{4}\log 2$

$ \Rightarrow $$4I = 2I - \frac{\pi }{4}\log 2$ [from Eq. (iii)]
therefore,$I = - \frac{\pi }{8}\log 2 = \frac{\pi }{8}\log \left( {\frac{1}{2}} \right)$
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